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Chuyển hết sang vế phải quy đồng ta được:
\(\frac{16x^2+4x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)}{6\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\frac{16x^2+8x^2+4x-48x^2+6x+1}{6\left(2x+1\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow24x^2-10x-1=0\Leftrightarrow\left(12x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{12}\\x=\frac{1}{2}\end{cases}}\)
a. (x + 3).(x2 - 1)
= x.x2 - x.1 + 3.x2 - 3.1
= x3 - x + 3x2 - 3
= x3 + 3x2 - x - 3
b. (3x + 2).(4x - 1)
= 3x.4x - 3x + 2.4x - 2
= 12x2 - 3x + 8x - 2
= 12x2 + 5x - 2
c. (2x - 3).(3x + 2)
= 2x.3x + 2x.2 - 3.3x - 3.2
= 6x2 + 4x - 9x - 6
= 6x2 - 5x - 6
d. (12x - 5).(4x + 1)
= 12x.4x + 12x - 5.4x - 5
= 48x2 + 12x - 20x - 5
= 48x2 - 8x - 5
e. (x - 3).(x2 + 3x + 9)
= x.x2 + x.3x + x.9 - 3x2 - 3.3x - 3.9
= x3 + 3x2 + 9x - 3x2 - 9x - 27
= x3 - 27 (Đây là dạng HĐT x3 - 33)
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
1.\(x^{16}-y^{16}=\left(x^8-y^8\right)\left(x^8+y^8\right)\)
2.\(x^3-125=x^3-5^3=\left(x-5\right)\left(x^2+5x+25\right)\)
\(-64+\frac{1}{8}x^3=\left(\frac{x}{2}\right)^3-4^3=\left(\frac{x}{2}-4\right)\left(\frac{x^2}{4}+2x+16\right)\)
\(8x^3+60x^2y+150xy^2+125y^3=\left(2x\right)^3+3.\left(2x\right)^2.\left(5y\right)+3.\left(2x\right).\left(5y\right)^2+\left(5y\right)^3\)
\(=\left(2x+5y\right)^3\)
Ta có : x2(x - 1)2 + x(x2 - 1) = 2(x + 1)2
<=> x2(x2 - 2x + 1) + x3 - x - 2(x2 + 2x + 1) = 0
<=> x4 - 2x3 + x2 + x3 - x - 2x2 - 4x - 2 = 0
<=> x4 - x3 - x2 - 5x - 2 = 0
?
Sửa đề: \(8x^3-\dfrac{1}{27}\)
\(=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{9}\right)\)
1) Ta có pt : \(4x^2+\frac{1}{x^2}=8x+\frac{4}{x}\)
\(\Leftrightarrow4x^2+4+\frac{1}{x^2}=8x+4+\frac{4}{x}\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2=4\left(2x+\frac{1}{x}\right)+4\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2-4\left(2x+\frac{1}{x}\right)+4=8\)
\(\Leftrightarrow\left(2x+\frac{1}{x}-2\right)^2=8\)
Đến đây dễ rồi nhé, chia 2 TH.
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn